Question

我一直在将程序MATLAB代码转换为OO C ++，到目前为止，这已经是相当不错的学习经验。我收到的错误是“ 0xC0000005：访问冲突写入位置0xFDFDFE05”，并且不确定这是什么意思。

以下是我的MATLAB代码，可以正常工作：

ms = 0;
for s = 1:Nx
    for m = 1:Ny
        ms = ms + 1;
        np = 0;
        for p = 1:Nx
        for n = 1:Ny
             np = np + 1;

             Xm = (m-0.5)*delta_x;
             Ys = (s-0.5)*delta_y;

             Xa = (n-1)*delta_x;
             Xb = n*delta_x;
             Ya = (p-1)*delta_y;
             Yb = p*delta_y;

             a  = (Yb-Ys);
             b  = (Xb-Xm);
             c  = (Ya-Ys);
             d  = (Xa-Xm);
             numA = b + sqrt(b.^2+a.^2);
             denA = d + sqrt(d.^2+a.^2);
             numB = a + sqrt(b.^2+a.^2);
             denB = c + sqrt(b.^2+c.^2);
             numC = d + sqrt(d.^2+c.^2);
             denC = b + sqrt(b.^2+c.^2);
             numD = c + sqrt(d.^2+c.^2);
             denD = a + sqrt(d.^2+a.^2);
             A = a.*log(numA./denA);
             B = b.*log(numB./denB);
             C = c.*log(numC./denC);
             D = d.*log(numD./denD);

             Z1(ms,np) = (A+B+(C+D))*k;
             end
         end
    end
end

以下是该函数，位于main.cpp之外的cpp文件中。（编辑：我进行了以下注释中建议的更改，并且现在可以按预期运行。）

double** zMatrix(double deltaX, double deltaY, double Nx, double Ny) {

double** Z = allocateMatrix(Nx*Nx, Ny*Ny);
int ms = 0;
for (int s = 0; s < Nx; s++)
{
    for (int m = 0; m < Ny; m++)
    {
        int np = 0;
        for (int p = 0; p < Nx; p++)
        {
            for (int n = 0; n < Ny; n++)
            {

                // Center of each observer patch
                double Xm = (m - 0.5) * deltaX;
                double Ys = (s - 0.5) * deltaY;

                // Source coordinates w/ corner (Xa, Ya) @ the origin (0, 0)
                // Lowest order basis function set:
                double Xa = (n - 1) * deltaX;
                double Xb = n * deltaX;
                double Ya = (p - 1) * deltaY;
                double Yb = p * deltaY;

        // Building analytical solution
                double a = (Yb - Ys);
                double b = (Xb - Xm);
                double c = (Ya - Ys);
                double d = (Xa - Xm);
                double numA = b + hypot(b, a); // sqrt(b*b + a * a);
                double denA = d + hypot(d, a); // sqrt(d*d + a * a);
                double numB = a + hypot(b, a); // sqrt(b*b + a * a);
                double denB = c + hypot(b, c); // sqrt(b*b + c * c);
                double numC = d + hypot(d, c); // sqrt(d*d + c * c);
                double denC = b + hypot(b, c); // sqrt(b*b + c * c);
                double numD = c + hypot(d, c); // sqrt(d*d + c * c);
                double denD = a + hypot(d, a); // sqrt(d*d + a * a);
                double A = a * log(numA / denA);
                double B = b * log(numB / denB);
                double C = c * log(numC / denC);
                double D = d * log(numD / denD);

                Z[ms][np] = (A + B + (C + D)) * k;
                np++;
            }
        }
        ms++;
    }
}
return Z;
}

这是我的分配矩阵函数：

double** allocateMatrix(double width, double height)
{
    // allocate the matrix
    double** zMatrix = new double*[width];
    for (int i = 0; i < width; i++)
    {
    zMatrix[i] = new double[height];
    }
    return zMatrix;
}

然后在main.cpp中，我这样称呼它：

double** Z = zMatrix(deltaX, deltaY, Nx, Ny);

我（通常）非常擅长调试，但实际上并没有给我太多帮助。

这里是运行时变量的捕获，我认为错误指向这里，即Z [ms]：

Answer 1

您正在将(width*height) x (width*height)元素写入width x height矩阵。因此，您的写作超出了范围。

我想您可以使用以下方法解决此问题：

double** Z = allocateMatrix(width*height, width*height);

双嵌套for循环中的访问冲突写入位置

1 个答案: